Calculating Energy Released When Hydrogen & Anti-Hydrogen Collide

[Mike12345]

So theirs this question

Particle accelerators can be used to accelerate particles up to 95% the speed of light. Determine the energy released when 1 hydrogen atom collides with an anti-hydrogen atom.

E=mc^2

= (1.00794)(2.998x10^8 x 0.95)^2
= 1.00794 x 8.1225 x 10^ 16
= 8.1869 x 10^16

I know this is wrong

When an anti hydrogen and hydrogen collide is the result 0?

 

[SteamKing]

The atomic weight of a hydrogen atom, 1.00794, should be converted into an actual mass figure, probably kg, in order to get E.

 

[eczeno]

you really need to make it clear what units you are using. you seem to be mixing them.

your calculation is wrong, but the energy released will not be zero.

the energy of an atom moving with (relativistic) momentum p and rest mass m is given by:

E^2 = (cp)^2 + (mc^2)^2

(note the c in mc^2 never changes, that is, you should never replace it with .95c or anything else)

and the same would be true for an anti-atom. so, in theory, you could get 2E out of such a collision, but certainly some of that will be converted back to mass very quickly if such a collision were to actually take place (that's what i think anyway).

 

[eczeno]

if you assume the collision happens at a slower (non-relativistic) speed, then the total energy released would be:

E = 2(mc^2)

hope this helps